(i) In a metre bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of `12.5 Omega`. Determine the resistance of X. Made of thick copper strips?
(ii). Determine the balance point of the bridge above if X and Y are interchanged.
(iii). what happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current?

Explanation

A metre bridge with resistors X and Y is represented in the given figure

(a) Balance point from end `A, I_(1) = 39.5` cm
Resistance of the resistor `Y = 12.5 Omega`
Condition for the balance is gives as,
`(X)/(Y)=(100-l_(1))/(l_(1))`
`X = (100-39.5)/(39.5) xx 12.5 = 8.2 Omega`
Therefore, the resistance of resistor X is `8.2 Omega`
The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula
(b) If X and Y are interchanged, then `l_(1) and 100-l_(1)` get interchanged
The balance point of the bridge will be `100 -l_(1)` from A
`100-l_(1)=100-39.5 = 60.5` cm
(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.